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9 RBTS PAPER

162390 Which of the following about polar molecule is true?
I. A polar molecule is one in which the centres of positive and negative charges are separated.
II. Polar molecule has no permanent dipole moment.
III. An ionic molecule such as $HCl$ are examples of polar molecules.

1 I only

2 II only

3 I and II

4 I and III

NCERT XII-I-65

9 RBTS PAPER

162391 If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor?

1 Because Electric current does not depends upon Drift velocity

2 Because the electron number density is small

3 Because the electron number density is enormous

4 None of these

NCERT XII-I-88

9 RBTS PAPER

162392 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5 Ω$ and at steam point is $5.23 Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795 Ω$ . Calculate the temperature of the bath.

1

300^{\circ} C

2

{200.56}^{\circ} C

3

{345.65}^{\circ} C

4

{445.65}^{\circ} C

Explanation:

$R_{0} = 5 Ω, R_{100} = 5.23 Ω$ and $R_{t} = 5.795 Ω$
Now, $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100, R_{t} = R_{0} (1 + α t)$
$= \frac{5.795 - 5}{5.23 - 5} \times 100 = \frac{0.795}{0.23} \times 100 = {345.65}^{\circ} C$

NCERT XII-I-92

9 RBTS PAPER

162393 The four arms of a Wheatstone bridge have the following resistances: $A B = 100 Ω, B C = 10 Ω$ , $C D = 5 Ω$ , and $D A = 60 Ω$ . A galvanometer of $15 Ω$ . resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across AC.

1

4.87 mA

.

2

10 mA

3

1 mA

4

9 mA

Explanation:

In mesh
$100 I_{1} + 15 I_{3} - 60 I_{2} = 0$
In mesh
$10 I_{1} - 10 I_{3} - 5 I_{2} - 5 I_{3} - 15 I_{3} = 0$
$10 I_{1} - 30 I_{3} - 5 I_{2} = 0$
From equation (i) and (ii) eliminate $I_{1}$
$315 I_{3} - 10 I_{2} = 0$
In mesh
$60 I_{2} + 5 I_{2} + 5 I_{3} = 10$
$5 I_{3} + 65 I_{2} = 10$
From equation (iii) and (iv) eliminate $I_{2}$
$I_{3} = 4.87 mA$

NCERT XII-I-103

9 RBTS PAPER

162390 Which of the following about polar molecule is true?
I. A polar molecule is one in which the centres of positive and negative charges are separated.
II. Polar molecule has no permanent dipole moment.
III. An ionic molecule such as $HCl$ are examples of polar molecules.

1 I only

2 II only

3 I and II

4 I and III

NCERT XII-I-65

9 RBTS PAPER

162391 If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor?

1 Because Electric current does not depends upon Drift velocity

2 Because the electron number density is small

3 Because the electron number density is enormous

4 None of these

NCERT XII-I-88

9 RBTS PAPER

162392 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5 Ω$ and at steam point is $5.23 Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795 Ω$ . Calculate the temperature of the bath.

1

300^{\circ} C

2

{200.56}^{\circ} C

3

{345.65}^{\circ} C

4

{445.65}^{\circ} C

Explanation:

$R_{0} = 5 Ω, R_{100} = 5.23 Ω$ and $R_{t} = 5.795 Ω$
Now, $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100, R_{t} = R_{0} (1 + α t)$
$= \frac{5.795 - 5}{5.23 - 5} \times 100 = \frac{0.795}{0.23} \times 100 = {345.65}^{\circ} C$

NCERT XII-I-92

9 RBTS PAPER

162393 The four arms of a Wheatstone bridge have the following resistances: $A B = 100 Ω, B C = 10 Ω$ , $C D = 5 Ω$ , and $D A = 60 Ω$ . A galvanometer of $15 Ω$ . resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across AC.

1

4.87 mA

.

2

10 mA

3

1 mA

4

9 mA

Explanation:

In mesh
$100 I_{1} + 15 I_{3} - 60 I_{2} = 0$
In mesh
$10 I_{1} - 10 I_{3} - 5 I_{2} - 5 I_{3} - 15 I_{3} = 0$
$10 I_{1} - 30 I_{3} - 5 I_{2} = 0$
From equation (i) and (ii) eliminate $I_{1}$
$315 I_{3} - 10 I_{2} = 0$
In mesh
$60 I_{2} + 5 I_{2} + 5 I_{3} = 10$
$5 I_{3} + 65 I_{2} = 10$
From equation (iii) and (iv) eliminate $I_{2}$
$I_{3} = 4.87 mA$

NCERT XII-I-103

9 RBTS PAPER

162390 Which of the following about polar molecule is true?
I. A polar molecule is one in which the centres of positive and negative charges are separated.
II. Polar molecule has no permanent dipole moment.
III. An ionic molecule such as $HCl$ are examples of polar molecules.

1 I only

2 II only

3 I and II

4 I and III

NCERT XII-I-65

9 RBTS PAPER

162391 If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor?

1 Because Electric current does not depends upon Drift velocity

2 Because the electron number density is small

3 Because the electron number density is enormous

4 None of these

NCERT XII-I-88

9 RBTS PAPER

162392 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5 Ω$ and at steam point is $5.23 Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795 Ω$ . Calculate the temperature of the bath.

1

300^{\circ} C

2

{200.56}^{\circ} C

3

{345.65}^{\circ} C

4

{445.65}^{\circ} C

Explanation:

$R_{0} = 5 Ω, R_{100} = 5.23 Ω$ and $R_{t} = 5.795 Ω$
Now, $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100, R_{t} = R_{0} (1 + α t)$
$= \frac{5.795 - 5}{5.23 - 5} \times 100 = \frac{0.795}{0.23} \times 100 = {345.65}^{\circ} C$

NCERT XII-I-92

9 RBTS PAPER

162393 The four arms of a Wheatstone bridge have the following resistances: $A B = 100 Ω, B C = 10 Ω$ , $C D = 5 Ω$ , and $D A = 60 Ω$ . A galvanometer of $15 Ω$ . resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across AC.

1

4.87 mA

.

2

10 mA

3

1 mA

4

9 mA

Explanation:

In mesh
$100 I_{1} + 15 I_{3} - 60 I_{2} = 0$
In mesh
$10 I_{1} - 10 I_{3} - 5 I_{2} - 5 I_{3} - 15 I_{3} = 0$
$10 I_{1} - 30 I_{3} - 5 I_{2} = 0$
From equation (i) and (ii) eliminate $I_{1}$
$315 I_{3} - 10 I_{2} = 0$
In mesh
$60 I_{2} + 5 I_{2} + 5 I_{3} = 10$
$5 I_{3} + 65 I_{2} = 10$
From equation (iii) and (iv) eliminate $I_{2}$
$I_{3} = 4.87 mA$

NCERT XII-I-103

9 RBTS PAPER

162390 Which of the following about polar molecule is true?
I. A polar molecule is one in which the centres of positive and negative charges are separated.
II. Polar molecule has no permanent dipole moment.
III. An ionic molecule such as $HCl$ are examples of polar molecules.

1 I only

2 II only

3 I and II

4 I and III

NCERT XII-I-65

9 RBTS PAPER

162391 If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor?

1 Because Electric current does not depends upon Drift velocity

2 Because the electron number density is small

3 Because the electron number density is enormous

4 None of these

NCERT XII-I-88

9 RBTS PAPER

162392 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5 Ω$ and at steam point is $5.23 Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795 Ω$ . Calculate the temperature of the bath.

1

300^{\circ} C

2

{200.56}^{\circ} C

3

{345.65}^{\circ} C

4

{445.65}^{\circ} C

Explanation:

$R_{0} = 5 Ω, R_{100} = 5.23 Ω$ and $R_{t} = 5.795 Ω$
Now, $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100, R_{t} = R_{0} (1 + α t)$
$= \frac{5.795 - 5}{5.23 - 5} \times 100 = \frac{0.795}{0.23} \times 100 = {345.65}^{\circ} C$

NCERT XII-I-92

9 RBTS PAPER

162393 The four arms of a Wheatstone bridge have the following resistances: $A B = 100 Ω, B C = 10 Ω$ , $C D = 5 Ω$ , and $D A = 60 Ω$ . A galvanometer of $15 Ω$ . resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across AC.

1

4.87 mA

.

2

10 mA

3

1 mA

4

9 mA

Explanation:

In mesh
$100 I_{1} + 15 I_{3} - 60 I_{2} = 0$
In mesh
$10 I_{1} - 10 I_{3} - 5 I_{2} - 5 I_{3} - 15 I_{3} = 0$
$10 I_{1} - 30 I_{3} - 5 I_{2} = 0$
From equation (i) and (ii) eliminate $I_{1}$
$315 I_{3} - 10 I_{2} = 0$
In mesh
$60 I_{2} + 5 I_{2} + 5 I_{3} = 10$
$5 I_{3} + 65 I_{2} = 10$
From equation (iii) and (iv) eliminate $I_{2}$
$I_{3} = 4.87 mA$

NCERT XII-I-103

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